Differentiation using 1st principles?

Question

find f'(x) by finding lim h->0 f(x+h)-f(x) over h. for:
f(x)=3x^2+4x+3

Answer 1

f(x) = 3x^2 + 4x + 3, the first pricinple for differentiation is lim h---->0 f(x+h)-f(x)/h, where 'h' means the small increment in 'x'.

f(x+h) = 3 (x+h)^2 + 4(x+h) +3

f(x+h) =3x^2 + 3h^2 + 6xh + 4x + 4h+ 3

f(x+h) - f(x) =3h^2 +6xh + 4h

f'(x) = lim h--->0 f(x+h) - f(x)/ h

f'(x) = lim h---> 0 3h^2 + 6xh +4h/h

f'(x) = lim h-->0 3h + 6x +4

when x approaches 0, but notice that 'h' is not equal to zero.

f'(x)= 6x + 4

Answer 2

The first thing we do is to evaluate f(x + h) - f(x). By the definition of f(x):

f(x + h) = 3(x + h)^2 + 4(x + h) + 3 = 3x^2 + 6xh + 3h^2 + 4x + 4h + 3
=> f(x + h) - f(x) = 3x^2 + 6xh + 3h^2 + 4x + 4h + 3 - 3x^2 - 4x - 3 = 6xh + 3h^2 + 4h

Now, dividing by h, we get 6x + 3h + 4. The limit of this expression as h goes to 0 is lim[h -> 0] (6x + 3h + 4) = 6x + 4.

Answer 3

If f(x) = 3x^2 + 4x + 3 then
f(x + h) = 3 (x + h)^2 + 4(x + h) + 3 = 3x^2 + 6hx + h^2 + 4x + 4h + 3
and f(x + h) - f(x) = 6hx + 4h + h^2 and (f(x + h) - f(x)) / h = (6hx + 4h + h^2) / h = 6hx / h + 4h / h + h^2 / h= 6x + 4 + h. Thus f'(x) = lim h->0 (f(x + h) - f(x))/h = lim h->0 (6x + 4 + h) = 6x + 4.+ 0 = 6x + 4.

Answer 4

f(x+h)-f(x) = 3(x+h)^2 + 4(x+h) + 3 - 3x^2 - 4x - 3
= 3x^2 + 6xh + 3h^2 + 4x + 4h + 3 - 3x^2 - 4x - 3
= 6xh +3h^2 + 4h

So...

lim(h->0) {f(x+h)-f(x)}/h = lim(h->0) {6xh+3h^2+4h}/h
= lim(h->0) 6x + 3h +4
= 6x+4

Answer 5

The first principle of differentiation is limits. Basically, by analyzing the slope between two points on a curve and making those two points approach one another, the limit condition just before they contact one another is the tangent slope of the curve. That is what the limit of f(x+h)-f(x) refers to, as h approaches zero.

I will not differentiate the particular equation you have supplied; do some more research and practice the technique and soon it will be easy.

Good luck!

Answer 6

plug it in...

f(x + h) = 3 * (x² + 2xh + h²) + 4(x+h) + 3
f(x) = 3x² + 4x +3

thus f(x+h)-f(x) = 6xh + h² + 4h

divide that by h = 6x + h + 4

let h ---> 0 and then f'(x) = 6x + 4

check to see if its right f'(x) = 6x + 4 yepo